1.

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere , both having the same mass and radius . The cylinder is free to rotate about its standard axis of symmetry , and the sphere is free to rotate about an axis passing through its centre . Which of the two will acquire a greater angular speed after a given time .

Answer»

Solution :Let M and R be mass and radius of the hollow cylinder and the solid SPHERE , then
Moment of INERTIA of the hollow cylinder about its axis of SYMMETRY , `I_(1) = MR^(2)`
Moment of inertia of the solid sphere about its axis through its centre `I_(2) = (2)/(5) MR^(2)`
Torque `TAU = I alpha`
where `alpha` is angular acceleration .
`I_(1) alpha_(1) = I_(2) alpha_(2)`
`(alpha_(2))/(alpha_(1)) = (I_(1))/(I_(2)) = (MR^(2))/((2)/(5) MR^(2)) = (5)/(2)`
`alpha_(2) = 2.5 alpha_(1) or alpha_(2) gt alpha_(1)`
From `omega = omega_(0) + alpha t` , we find that for given `omega_(0)` and `t , omega_(2) gt omega_(1)` , angular speed of solid sphere will be greater than the angular speed of hollow cylinder .


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